Homework questions – Week 11

(1)

Homework questions – Week 11

Jean-Baptiste Campesato April 5^th, 2021 to April 9^th, 2021

Exercise 1.

Given three sets𝐸, 𝐹 , 𝐺, prove that if𝐸 ⊂ 𝐹 ⊂ 𝐺and|𝐸| = |𝐺|then|𝐸| = |𝐹 |.

Exercise 2.

Given a set𝑆, prove that|𝒫(𝑆)| = |{0, 1}^𝑆|where{0, 1}^𝑆denotes the set of functions𝑆 → {0, 1}.

Remark: this formula generalizes the fact that if𝑆is a finite set with𝑛 = |𝑆|then|𝒫(𝑆)| = 2^𝑛. Therefore it is common to denote the powerset of a set𝑆by2^𝑆 ≔ 𝒫(𝑆).

Exercise 3.

1. What is|{0, 1}^ℕ|? i.e. what is the cardinality of the set of functionsℕ → {0, 1}?

2. What is|ℕ^{0,1}|? i.e. what is the cardinality of the set of functions{0, 1} → ℕ?

Exercise 4.

1. What is the cardinality of𝑆 = {𝐴 ∈ 𝒫(ℕ) ∶ 𝐴is finite}.

2. Is𝑇 = {𝐴 ∈ 𝒫(ℕ) ∶ 𝐴is infinite}countable?

Exercise 5.

Prove that any set𝑋of pairwise disjoint intervals which are non-empty and not reduced to a singleton is countable,

i.e. if𝑋 ⊂ 𝒫(ℝ)satisfies

(i) ∀𝐼 ∈ 𝑋, 𝐼is an interval which is non-empty and not reduced to a singleton (ii) ∀𝐼, 𝐽 ∈ 𝑋, 𝐼 ≠ 𝐽 ⟹ 𝐼 ∩ 𝐽 = ∅

then𝑋is countable.

Exercise 6.

Prove that a set is infinite if and only if it admits a proper subset of same cardinality.

Exercise 7.

1. Prove thatℝ ⧵ ℚis not countable.

2. Prove that|ℝ ⧵ ℚ| = |ℝ|.

Exercise 8.

Prove that|(0, 1)| = |ℝ|.

Exercise 9.

1. Prove that|ℝ²| = |ℝ|.

2. Prove that∀𝑛 ∈ ℕ ⧵ {0}, |ℝ^𝑛| = |ℝ|.

3. Prove that|ℝ^ℕ| = |ℝ|whereℝ^ℕis the set of sequences/functionsℕ → ℝ.

Exercise 10.

Set𝑆² ≔ {(𝑥, 𝑦, 𝑧) ∈ ℝ³ ∶ 𝑥²+ 𝑦²+ 𝑧² = 1}. Prove that|𝑆²| = |ℝ|.

Exercise 11.

What is the cardinality of the set𝑆of all circles in the plane?

(2)

Cheatsheet

Recollection of some results about cardinality

Definition. We say that two sets 𝐸 and𝐹 have samecardinality, denoted by|𝐸| = |𝐹 |, if there exists a bijection𝑓 ∶ 𝐸 → 𝐹.

Proposition.

1. If𝐸is a set then|𝐸| = |𝐸|.

2. Given two sets𝐸and𝐹, if|𝐸| = |𝐹 |then|𝐹 | = |𝐸|.

3. Given three sets𝐸,𝐹 and𝐺, if|𝐸| = |𝐹 |and|𝐹 | = |𝐺|then|𝐸| = |𝐺|.

Theorem. A set𝐸is infinite if and only if for every𝑛 ∈ ℕthere exists𝑆 ⊂ 𝐸such that|𝑆| = 𝑛.

Definition. Given two sets𝐸and𝐹, we write|𝐸| ≤ |𝐹 |if there exists an injective function𝑓 ∶ 𝐸 → 𝐹. Proposition.

1. If𝐸is a set then|𝐸| ≤ |𝐸|.

2. Given two sets𝐸and𝐹, if|𝐸| ≤ |𝐹 |and|𝐹 | ≤ |𝐸|then|𝐸| = |𝐹 |Cantor–Schröder–Bernstein theorem.

3. Given three sets𝐸,𝐹 and𝐺, if|𝐸| ≤ |𝐹 |and|𝐹 | ≤ |𝐺|then|𝐸| ≤ |𝐺|. Proposition. If𝐸 ⊂ 𝐹 then|𝐸| ≤ |𝐹 |.

Proposition. If|𝐸₁| = |𝐸₂|and|𝐹₁| = |𝐹₂|then|𝐸₁× 𝐹₁| = |𝐸₂× 𝐹₂|.

Theorem. Given two sets𝐸and𝐹,|𝐸| ≤ |𝐹 |if and only if there exists a surjective function𝑔 ∶ 𝐹 → 𝐸.

Theorem. Given two sets𝐸and𝐹, if|𝐸| = |𝐹 |then|𝒫(𝐸)| = |𝒫(𝐹 )|.

Notation. We setℵ₀ ≔ |ℕ|(pronouncedaleph nought).

Definition. A set𝐸is countable if either𝐸is finite or|𝐸| = ℵ₀. Proposition. If𝑆 ⊂ ℕis infinite then|𝑆| = ℵ₀.

Proposition. A set𝐸is countable if and only if|𝐸| ≤ ℵ₀(i.e. there exists an injection𝑓 ∶ 𝐸 → ℕ), otherwise stated𝐸is countable if and only if there exists a bijection between𝐸and a subset ofℕ.

Proposition. |ℕ × ℕ| = ℵ₀

Theorem. A countable union of countable sets is countable,

i.e. if𝐼is countable and if for every𝑖 ∈ 𝐼,𝐸_𝑖is countable then⋃_𝑖∈𝐼𝐸_𝑖is countable.

Theorem. If𝐸is an infinite set then there exists𝑇 ⊂ 𝐸such that|𝑇 | = ℵ₀, i.e.ℵ₀is the least infinite cardinal.

Theorem. |ℤ| = ℵ₀ Theorem. |ℚ| = ℵ₀ Theorem. ℵ₀< |ℝ|

Theorem(Cantor’s theorem). Given a set𝐸,|𝐸| < |𝒫(𝐸)|.

Theorem. |ℝ| = |𝒫(ℕ)|

(3)

Sample solutions to Exercise 1.

Since𝐸 ⊂ 𝐹, we know that|𝐸| ≤ |𝐹 |.

Besides, since𝐹 ⊂ 𝐺, we have|𝐹 | ≤ |𝐺| = |𝐸|.

By Cantor–Schröder–Bernstein theorem, we have|𝐸| = |𝐹 |.

We define𝜓 ∶ 𝒫(𝑆) → {0, 1}^𝑆 by𝜓(𝐴)(𝑥) = {

1 if𝑥 ∈ 𝐴 0 otherwise . Let’s prove that𝜓 is a bijection:

• 𝜓is injective.

Let𝐴, 𝐵 ⊂ 𝑆be such that𝐴 ≠ 𝐵.

WLOG we may assume that there exists𝑥 ∈ 𝑆such that𝑥 ∈ 𝐴and𝑥 ∉ 𝐵.

Therefore𝜓(𝐴)(𝑥) = 1and𝜓(𝐵)(𝑥) = 0. Thus𝜓(𝐴) ≠ 𝜓(𝐵).

• 𝜓is surjective.

Let𝑓 ∶ 𝑆 → {0, 1}be a function. Define𝐴 = {𝑥 ∈ 𝑆 ∶ 𝑓 (𝑥) = 1}. Then𝑓 = 𝜓(𝐴).

Therefore|𝒫(𝑆)| = |{0, 1}^𝑆|. Sample solutions to Exercise 3.

1. |{0, 1}^ℕ| = |𝒫(ℕ)| = |ℝ|

2. The idea here is that a function{0, 1} → ℕis characterized by the values of0and1.

Define𝜓 ∶ ℕ^{0,1}→ ℕ × ℕby𝜓(𝑓 ) = (𝑓 (0), 𝑓 (1)).

• 𝜓 is injective: let𝑓 , 𝑔 ∶ ℕ → {0, 1}be such that𝜓(𝑓 ) = 𝜓(𝑔). Then(𝑓 (0), 𝑓 (1)) = (𝑔(0), 𝑔(1))so that𝑓 (0) = 𝑔(0)and𝑓 (1) = 𝑔(1). Therefore𝑓 = 𝑔.

• 𝜓 is surjective: let(𝑎, 𝑏) ∈ ℕ × ℕ. Define 𝑓 ∶ {0, 1} → ℕ by𝑓 (0) = 1 and𝑓 (1) = 𝑏. Then 𝜓(𝑓 ) = (𝑎, 𝑏).

Therefore|ℕ^{0,1}| = |ℕ × ℕ| = ℵ0.

1. Define𝑓 ∶ 𝑆 → ℕby𝑓 (𝐴) = ∑

𝑘∈𝐴

2^𝑘.

Then𝑓 is bijective by existence and uniqueness of the binary positional representation of a natural number. Therefore|𝑆| = |ℕ| = ℵ₀.

2. Assume that𝑇 is countable then𝒫(ℕ) = 𝑆 ⊔ 𝑇 is countable as the union of countable sets.

Which is a contradiction since|𝒫(ℕ)| = |ℝ| > ℵ₀.

Let𝑋be as in the statement.

For𝐼 ∈ 𝑋, we can find𝑞_𝐼 ∈ 𝐼 ∩ ℚsince𝐼 is an interval which is non-empty and not reduced to a singleton.

Define𝑓 ∶ 𝑋 → ℚby𝑓 (𝐼) = 𝑞_𝐼. Let’s prove that𝑓 is injective.

Let𝐼, 𝐽 ∈ 𝑋 such that𝑞 ≔ 𝑓 (𝐼) = 𝑓 (𝐽 ). Then𝑞 = 𝑞_𝐼 ∈ 𝐼 and𝑞 = 𝑞_𝐽 ∈ 𝐽. Therefore𝑞 ∈ 𝐼 ∩ 𝐽 ≠ ∅. Thus 𝐼 = 𝐽 (use the contrapositive of (ii)).

Hence|𝑋| ≤ |ℚ| = ℵ₀. So𝑋is countable.

(4)

⇒Let’s prove that any infinite set admits a proper subset of same cardinality.

Let𝑋be an infinite set. We want to construct𝑆 ⊊ 𝑋satisfying|𝑆| = |𝑋|.

Since𝑋is infinite,ℵ₀ ≤ |𝑋|, i.e. there exists an injective function𝑓 ∶ ℕ → 𝑋.

We define𝑔 ∶⎧⎪

⎨⎪

⎩

𝑋 → 𝑋

𝑥 ↦ 𝑓 (𝑛 + 1) if∃𝑛 ∈ ℕ, 𝑥 = 𝑓 (𝑛)

𝑥 ↦ 𝑥 if𝑥 ∉Im(𝑓 )

.

• 𝑔is well-defined: given𝑥 ∈ 𝑋, if∃𝑛, 𝑚 ∈ ℕ, 𝑥 = 𝑓 (𝑛) = 𝑓 (𝑚)then𝑛 = 𝑚since𝑓 is injective.

• 𝑔is injective: let𝑥, 𝑦 ∈ 𝑋 be such that𝑔(𝑥) = 𝑔(𝑦).

– First case: 𝑔(𝑥) = 𝑔(𝑦) ∈ Im(𝑓 ) then there exists𝑛, 𝑚 ∈ ℕ such that 𝑥 = 𝑓 (𝑛) and𝑦 = 𝑓 (𝑚).

Since𝑓 (𝑛 + 1) = 𝑔(𝑥) = 𝑔(𝑦) = 𝑓 (𝑚 + 1), we get that𝑛 = 𝑚by injectiveness of 𝑓. Therefore 𝑥 = 𝑓 (𝑛) = 𝑓 (𝑚) = 𝑦.

– Second case:𝑔(𝑥) = 𝑔(𝑦) ∉Im(𝑓 )then𝑥 = 𝑔(𝑥) = 𝑔(𝑦) = 𝑦.

Note that𝑓 (0) ∉Im(𝑔), thus𝑓 (0) ∈ 𝑋 ⧵Im(𝑔). Besides𝑔 ∶ 𝑋 →Im(𝑔)is a bijection. Hence𝑆 =Im(𝑔) satisfies𝑆 ⊊ 𝑋and|𝑋| = |𝑆|.

⇐We are going to prove the contrapositive: if a set is finite then it doesn’t admit a proper subset of same cardinality.

Let𝑋be a finite set. Let𝑆 ⊊ 𝑋be a proper subset.

Then there exists𝑥₀ ∈ 𝑋 ⧵ 𝑆so that𝑆 ⊔ {𝑥₀} ⊂ 𝑋and hence|𝑆 ⊔ {𝑥₀}| = |𝑆| + 1 ≤ |𝐸|, i.e.|𝑆| < |𝐸|.

1. Assume by contradiction thatℝ ⧵ ℚis countable. Thenℝ = (ℝ ⧵ ℚ) ∪ ℚis countable as the union of two countable sets. Hence a contradiction.

2. One way to solve this question is to take an injective functionℝ → ℝwhose range is a proper interval ofℝand then to move the rational values in the complement of the range after making them irrational.

For instance:

Define𝑓 ∶ ℝ → ℝ ⧵ ℚby

𝑓 (𝑥) = {

𝑒^𝑥 if𝑒^𝑥∉ ℚ

−𝑒^𝑥− 𝑒 otherwise

• 𝑓 is well-defined: if𝑒^𝑥∈ ℚthen−𝑒^𝑥− 𝑒 ∈ ℝ ⧵ ℚ(since−𝑒^𝑥∈ ℚand−𝑒 ∈ ℝ ⧵ ℚ).

• 𝑓 is injective: let𝑥, 𝑦 ∈ ℝbe such that𝑓 (𝑥) = 𝑓 (𝑦).

– First case: 𝑓 (𝑥) = 𝑓 (𝑦) > 0then𝑓 (𝑥) = 𝑒^𝑥and𝑓 (𝑦) = 𝑒^𝑦thus𝑒^𝑥= 𝑓 (𝑥) = 𝑓 (𝑦) = 𝑒^𝑦and then 𝑥 = 𝑦since exp is injective.

– Second case: 𝑓 (𝑥) = 𝑓 (𝑦) < 0then𝑓 (𝑥) = −𝑒^𝑥− 𝑒and𝑓 (𝑦) = −𝑒^𝑦− 𝑒thus−𝑒^𝑥− 𝑒 = 𝑓 (𝑥) = 𝑓 (𝑦) = −𝑒^𝑦− 𝑒, so that𝑒^𝑥= 𝑒^𝑦and hence𝑥 = 𝑦since exp is injective.

Note that𝑓 (𝑥) = 𝑓 (𝑦) ≠ 0since0 ∈ ℚ.

Thus|ℝ| ≤ |ℝ ⧵ ℚ|. Besides|ℝ ⧵ ℚ| ≤ |ℝ|sinceℝ ⧵ ℚ ⊂ ℝ.

Hence|ℝ| = |ℝ ⧵ ℚ|by Cantor–Schröder–Bernstein theorem.

Comment: (using the axiom of choice) it is true that if𝐴and𝐵 are infinite sets then|𝐴 ∪ 𝐵| =max(|𝐴|, |𝐵|) (but this statement was not proved in class, so you can’t use it).

Therefore, since|ℚ| < |ℝ ⧵ ℚ|,|ℝ| = |(ℝ ⧵ ℚ) ∪ ℚ| =max(|ℝ ⧵ ℚ|, |ℚ|) = |ℝ ⧵ ℚ|.

(5)

Define𝑓 ∶ ℝ → (0, 1)by𝑓 (𝑥) = ^{arctan(𝑥)+}

𝜋 2

𝜋 . Then

• 𝑓 is well-defined:

for𝑥 ∈ ℝ,−^𝜋

2 <arctan(𝑥) < ^𝜋

2 thus0 <arctan(𝑥)+^𝜋

2 < 𝜋and hence0 < ^{arctan(𝑥)+}

𝜋 2

𝜋 < 1, i.e.𝑓 (𝑥) ∈ (0, 1).

• 𝑓 is bijective:prove it using thatarctan∶ ℝ → (−^𝜋2,^𝜋

2)is bijective.

Therefore|(0, 1)| = |ℝ|.

There are lots of such bijections, for instance:

(0, 1) → ℝ

𝑥 ↦ ¹

1+𝑒^𝑥

(0, 1) → ℝ 𝑥 ↦ ^2𝑥−1

𝑥−𝑥²

ℝ → (0, 1) 𝑥 ↦ 𝑒^−𝑒^𝑥

1. First method:

We define𝑓 ∶ (0, 1) × (0, 1) → (0, 1)as follows. Let(𝑥, 𝑦) ∈ (0, 1) × (0, 1).

Denote the proper decimal expansions of𝑥and𝑦by𝑥 =

+∞

∑𝑘=1

𝑎_𝑘10^−𝑘= 0.𝑎₁𝑎₂…where𝑎_𝑘∈ {0, 1, … , 9}

are not all equal to0and𝑦 =

+∞

∑𝑘=1

𝑏_𝑘10^−𝑘= 0.𝑏₁𝑏₂…similarly.

Then we set𝑓 (𝑥, 𝑦) =

+∞

∑𝑘=0

𝑎_𝑘10^−(2𝑘+1)+

+∞

∑𝑘=1

𝑏_𝑘10^−2𝑘 = 0.𝑎₁𝑏₁𝑎₂𝑏₂… =

+∞

∑𝑘=1

𝑐_𝑘10^−𝑘where

𝑐_𝑘= {

𝑎_𝑛 if∃𝑛 ∈ ℕ ⧵ {0}, 𝑘 = 2𝑛 𝑏_𝑛 if∃𝑛 ∈ ℕ, 𝑘 = 2𝑛 + 1

Then𝑓 a bijection by existence and uniqueness of the proper decimal expansion.

Hence|(0, 1) × (0, 1)| = |(0, 1)|.

Since|(0, 1)| = |ℝ, we get|ℝ × ℝ| = |(0, 1) × (0, 1)| = |(0, 1)| = |ℝ|.

Second method:

Define𝑓 ∶ 𝒫(ℕ) × 𝒫(ℕ) → 𝒫(ℕ)by𝑓 (𝐴, 𝐵) = {2𝑘 ∶ 𝑘 ∈ 𝐴} ∪ {2𝑙 + 1 ∶ 𝑙 ∈ 𝐵}.

Then𝑓 is bijective(prove it).

Thus|𝒫(ℕ) × 𝒫(ℕ)| = |𝒫(ℕ)|.

Since|ℝ| = |𝒫(ℕ)|, we get|ℝ × ℝ| = |𝒫(ℕ) × 𝒫(ℕ)| = |𝒫(ℕ)| = |ℝ|.

2. Let’s prove by induction on𝑛 ∈ ℕ ⧵ {0}that|ℝ^𝑛| = |ℝ|.

• Base case at𝑛 = 1:thenℝ¹ = ℝthus|ℝ¹| = |ℝ|.

• Inductive step:assume that|ℝ^𝑛| = |ℝ|for some𝑛 ∈ ℕ ⧵ {0}. Then

|ℝ^𝑛+1| = |ℝ^𝑛× ℝ|

= |ℝ × ℝ| since|ℝ^𝑛| = |ℝ|and|ℝ| = |ℝ|

= |ℝ| by the previous question

3. One idea here is to notice that|ℝ^ℕ| = |({0, 1}^ℕ)^ℕ| = |{0, 1}^ℕ×ℕ| = |{0, 1}^ℕ| = |ℝ|. Since we have not covered arithmetic of cardinals, we need to prove each equality.

• Since|ℝ| = |𝒫(ℕ)| = |{0, 1}^ℕ|, there exists a bijection𝜓 ∶ ℝ → {0, 1}^ℕ. We define𝜑 ∶ ℝ^ℕ→ ({0, 1}^ℕ)^ℕby𝜑(𝑓 ) = 𝜓 ∘ 𝑓 ∶ ℕ → {0, 1}^ℕ.

Then𝜑is a bijection(check it), and thus|ℝ^ℕ| = |({0, 1}^ℕ)^ℕ|.

(6)

• We define𝜉 ∶ {0, 1}^ℕ×ℕ → ({0, 1}^ℕ)^ℕby𝜉(𝑓 ) ∶ {

ℕ → {0, 1}^ℕ 𝑛 ↦ (𝑚 ↦ 𝑓 (𝑛, 𝑚)) . Check that𝜉is a bijection. Therefore|({0, 1}^ℕ)^ℕ| = |{0, 1}^ℕ×ℕ|.

• Since|ℕ × ℕ| = |ℕ|, there exists a bijection𝜁 ∶ ℕ × ℕ → ℕ.

We define𝛾 ∶ {0, 1}^ℕ→ {0, 1}^ℕ×ℕby𝛾(𝑓 ) = 𝑓 ∘ 𝜁.

Check that𝛾is a bijection. Therefore|{0, 1}^ℕ×ℕ| = |{0, 1}^ℕ|.

Define𝑓 ∶ (0, 1) → 𝑆²by𝑓 (𝑡) = (cos𝑡,sin𝑡, 0). Then𝑓 is well-defined and injective.

Thus|ℝ| = |(0, 1)| ≤ |𝑆²|.

Besides, since𝑆² ⊂ ℝ³, we have that|𝑆²| ≤ |ℝ³| = |ℝ|.

By Cantor–Schröder–Bernstein theorem, we get that|𝑆²| = |ℝ|.

A circle is characterized by its center and its radius. Therefore there is a bijectionℝ²× (0, +∞) → 𝑆 mapping (𝑥, 𝑦, 𝑟)to the circle centered at(𝑥, 𝑦)of radius𝑟.

Thus|𝑆| = |ℝ²× (0, +∞)|.

Since exp∶ ℝ → (0, +∞)is a bijection, we have|(0, +∞)| = |ℝ|. Hence|ℝ²× (0, +∞)| = |ℝ³| = |ℝ|.

Therefore|𝑆| = |ℝ|.